Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction

Simon777

1. The problem statement, all variables and given/known data
Two horses pull horizontally on ropes attached to a tree stump. Each horse pulls with a force of magnitude F. If the resultant force (vector R) has the magnitude R = 1.51 F, what is the angle (in degrees) between the two ropes?


2. Relevant equations



3. The attempt at a solution
Ry= 0+cos
Rx= 1+ sin

R= 1.51

sin=cos/1.54
tan= 1/1.54
angle= 33 degrees

33 * 2 = total angle for both of 66 degrees

Simon777

I'm on my 5th hour for this problem and still can't get it. I realize it's necessary to assign an arbitrary value of say 1 to F. That gives F=1 and R=1.51, but everything I've tried to do with that information has failed.

Allenman

Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G.

So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1).

F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees

That's just half of your angle though since you split it into two triangles
49.025*2 = 98.05 degrees

Simon777

The professor gave an answer of 81.9 degrees.

Allenman

Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.

Simon777

I really appreciate the response, I initially did what you did and was told to rethink it and that it was 81.9. That's what's been making me go insane. I don't know why this method is wrong.

Simon777

You had it right and explained it right, but used arcsin instead of arccos. I get how to do this now, thank you so much for helping me finally understand this.